### Video Transcript

Given the polar curve defined by π is equal to π of π, form an expression for the slope of the curve the derivative of π¦ with respect to π₯ in terms of π and π.

The question gives us a polar curve defined by the equation π is equal to π of π. And itβs asking us to form an expression for the slope of the curve. Thatβs the derivative of π¦ with respect to π₯. And it wants this in terms of π and π. To do this, we recall the chain rule, which tells us that the derivative of π¦ with respect to π₯ is equal to the derivative of π¦ with respect to π multiplied by the derivative of π with respect to π₯. So, this is the first step. This gives us an equation for our slope function dπ¦ by dπ₯.

We now recall our polar equations, π¦ is equal to π sin π and π₯ is equal to π cos π. We recall the question is asking us to write our slope function in terms of only π and π. So, it would be a good idea to rewrite any equations in terms of π and π where possible. Luckily, we can use the substitution π is equal to π of π to rewrite these polar equations.

At this point, we could calculate the derivative of π¦ with respect to π without any problems. However, we canβt directly calculate the derivative of π with respect to π₯, since weβre given π₯ as a function of π. We can get around this by recalling that multiplying by the derivative of π with respect to π₯ is the same as dividing by the derivative of π₯ with respect to π. So, we can find an expression for our slope function dπ¦ by dπ₯ by first calculating dπ¦ by dπ and then dividing this by dπ₯ by dπ. So, to calculate the derivative of π¦ with respect to π, we differentiate π of π multiplied by sin π with respect to π.

We can differentiate this by recalling the product rule, which tells us the derivative of the product of two functions π’ and π£ is equal to π’ prime multiplied by π£ plus π£ prime multiplied by π’. So, we first differentiate our function π of π to get π prime of π and multiply this by sin π. Then, we differentiate sin π, which we recall is cos π. And we multiply this by π of π. We can then do the same to calculate the derivative of π₯ with respect to π. So, thatβs the derivative of π of π multiplied by the cos of π with respect to π.

We differentiate π of π to get π prime of π and then multiply this by the cos of π. And we add onto this the derivative of the cos of π, which we recall is negative sin π. And we multiply this by π of π. Weβre now ready to substitute this into our equation for our slope function. We have that dπ¦ by dπ₯ is equal to dπ¦ by dπ, divided by dπ₯ by dπ. Substituting the equation we got for the derivative of π¦ with respect to π gives us the numerator of π prime of π multiplied by sin π plus π of π multiplied by the cos of π.

And then, substituting the equation we got for the derivative of π₯ with respect to π gives us a denominator of π prime of π multiplied by the cos of π minus π of π multiplied by the sin of π. And so, we have found the formula for our slope function dπ¦ dπ₯ in terms of the function π and π. And itβs fine to just quote this. We donβt need to derive this every time. And so, in conclusion, we have shown that if we are given a polar curve defined by the equation π of π. Then, we can rewrite the slope function the derivative of π¦ with respect to π₯ as π prime of π multiplied by sin π plus π of π multiplied by the cos of π all divided by π prime of π multiplied by the cos of π minus π of π multiplied by the sin of π.